Procedure:

1. Consider a flat disk spinning counter-clockwise as shown below:

 

 

 

 

 

At each point shown on the disk draw an arrow to show the direction of

the velocity at that point.

 If the disk is rotating at 1 revolution every 5 seconds, how much time does it take point A to complete one circuit? Point B? Point C?

 If the disk has a radius of 0.25 m, can you rank the velocities of points A, B and C in increasing order?

 

 

 

 

 

     If point A and point D are the same distance from the center of the disk, draw velocity vectors accurately representing the velocity of each of these points.

 

How would the velocity vectors at point E and F appear?

 

Draw a conclusion about the velocity vectors and the radius of the points:

 

 

How many degrees does each point rotate through in 1 second?

	Point A	Point B	Point C	Point D	Point E	Point F
Degrees
Radians
Time
Radians/time

 

Calculate the “Radians/time” above for each point.  This quantity is called the “angular speed” when the angles are expressed in radians.  This expresses the magnitude of the wheels rotation.  To describe the “angular velocity” we need to express whether the wheel is rotating clockwise or counter-clockwise.  The positive direction is counter-clockwise.

 

Interim Discussion:

Describing rotational motion of an object, we must measure the angle through which a point moves about the axis of rotation and the time it takes to move through that angle.  This is analogous to measuring (x, t) in linear motion.  In the case of rotational motion we measure (q, t).  In the simplest terms, all of the equations that you learned relating to linear motion now can be applied to rotational motion by just replacing all the ‘x’s with ‘q’s.  The symbol for angular speed is omega, w, and has units of rad/s.

 

Problem: Electrical wire of diameter 0.75 cm is wound on a spool with a radius of 30 cm and a length of 24 cm.  (a) Through how many radians must the spool be turned to wrap one even layer of wire? (b) What is the length of this wound wire?

 

 

 

 

Increasing or Decreasing Rotational Speed.

The figure shows a rotating disk that is speeding up at a constant rate.  Point A is

shown at several times separated by 1 second.  The distance from the axis of

rotation to point A is 10 cm.

 Calculate the angular speed for each interval.

Points	A1	A2	A3	A4
Angular Speed

Calculate the change in angular speed between each interval.

Points	A1	A2	A3	A4
Angular Speed Change

 

Calculate the change in angular speed divided by the time difference for each interval.

Points	A1 to A2	A2 to A3	A3 to A4
Angular Speed Change/Time

This quantity, angular speed/time, is the angular acceleration, a, in units of radians/sec². 

 

Problem: The driver of a car sets the cruise control and ties the steering wheel so that the car travels at a uniform speed of 15 m/s in a circle with a diameter of 120 m.  (a) Through what angular distance does the car move in 4.0 minutes? (b) What arc distance does it travel in this time?

 Interim Discussion on Correspondence of Linear and Rotational Motion

 

We now have all the corresponding quantities defined for rotational motion, i.e. t, q, w, and a.  This completes our rotational kinematics definitions.  We can use all of our previous linear motion kinematics equations with the appropriate replacement of ‘t’, ‘x’, ‘v’ and ‘a’ with ‘t’, ‘q’ ‘w’, and ’a’.

 

Problem: A bicycle being repaired is turned upside down, and one wheel is rotated at a rate of 60 rpm.  If the wheel slows uniformly to a stop in 15 s, how many revolutions does it make during this time?

 

Centripetal Force

Just as there were simple kinematics expressions for circular motion that were easily derived from linear kinematics, so, too, is there a analog for Newton’s Second Law.  First, let’s recall Newton’s First and Second Laws.  The first law informed us that if we see an object moving at a constant velocity (which means constant speed and constant direction) then we must conclude that no force is acting on the object under consideration.  Conversely, if a force is present then the object’s speed and/or direction will change.

 

Let’s deal with the situation where the speed is constant, the direction is changing as the object moves about in a circle.  Even if we can’t see the reason for the circular motion, we must conclude that a force is acting toward the center of the circle that keeps the object on its circular path.  Since there is clearly a force act work here Newton’s second law must apply.  We must modify the usual form of the 2^nd law since the speed is constant while the direction is changing.  A mathematical derivation will show that the normal linear acceleration vector, a, is replaced by the centripetal acceleration of magnitude, a_c = v²/r, directed toward the center of the circular path.  The word ‘centripetal’ actually means ‘centrally directed’.  By replacing the linear acceleration with the centripetal acceleration, the 2^nd law expression becomes,

This expression is a mathematical statement of a physical relationship.  It expresses the equality between the description of circular motion of radius, r, and speed, v, to the force, F_c, responsible for the motion.

 

Example: A stone is tied to a 0.5 m string and whirled at a constant speed of 4.0 m/s in a vertical circle.  What is the magnitude of its acceleration in m/s² at the bottom of the circle?

 

Solution: r = 0.5 m and v = 4.0 m/s,

               a_c = v²/r = (4.0 m/s)²/0.5 m

                             = 32 m/s². This acceleration is directed upward.

 

 

 

 

Example: What is the centripetal force of the Earth on the Moon?  Relevant data would be:  Earth-Moon distance = 362,000 km, orbital period = 28 days, Moon’s mass = 7.4 x 10²² kg.

 

Solution: r = 362,000 km = 362,000,000 m.

Mass, m = 7.4 x 10²² kg.

 

To find centripetal force, solve F_c = mv²/r.  Need to find v.  Several methods can be used here:

  v = 2pr/T = 2*3.14*362,000,000 m/(28 days*24 h/day*3600 s/h)

  v =  940 m/s.

 

Now solve for the centripetal force:

F_c = 7.4 x 10²² kg (940 m/s)²/(362,000,000 m) = 1.8 x 10²⁰ N. 

That may seem like a lot of force, but remember it is acting on a very large object.
 

 

 

 

Centripetal Force Problems.

 

 

1.       A car is traveling 25 m/s around a level curve of radius 120 m.  What is the minimum value of the coefficient of static friction between the tires and the road needed to prevent skidding?

 

 

 

 

2. A huge pendulum consists of a 200 kg ball at the end of 15 m of cable.  If the pendulum is drawn back to 37^o and released, what is the maximum force on the cable as it swings back and forth?

 

 

 

 

 

 

 

 

Problem: A pickup of 4.25 m length and mass of 1500 kg attempts to jump a ravine that is 10 m wide and the opposite side is 2.96 m below the launch side.  The driver makes a wide circular arc of radius 0.333 km and approaches the ravine.  What is the minimum centripetal acceleration of the truck in order to clear the ravine?

 

 

 

Problem: A car on a circular track of radius 0.30 km accelerates from rest with a constant angular acceleration of 0.005 rad/s².  How much time does it take to complete one lap?  What is the total acceleration of the car when it has reached one-half a lap?

 

 

Problem: Compare the gravitational attraction of the Sun on the Earth with the Moon on the Earth.  The masses and distances for the Sun and Moon are M_sun = 2 x 10³⁰kg, M_moon = 7.4 x 10²² kg, Earth-Sun distance= 1.5 x 10¹¹ m, Earth-Moon distance = 3.8 x 10⁸ m.   G = 6.67 x 10^-11 Nm²/kg².